Solutions — Product and Quotient Rules for All Function Types

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1. Show Solution
   y = x² sin x
   Let u = x², v = sin x
   u′ = 2x, v′ = cos x
   Product rule: dy/dx = u′v + uv′
   dy/dx = 2x sin x + x² cos x
2. Show Solution
   y = e^x ln x
   Let u = e^x, v = ln x
   u′ = e^x, v′ = 1/x
   dy/dx = e^x ln x + e^x · (1/x)
   dy/dx = e^x(ln x + 1/x)
3. Show Solution
   y = sin x / x
   Let u = sin x, v = x
   u′ = cos x, v′ = 1
   Quotient rule: dy/dx = (u′v − uv′) / v²
   dy/dx = (cos x · x − sin x · 1) / x²
   dy/dx = (x cos x − sin x) / x²
4. Show Solution
   y = x³ / e^x
   Let u = x³, v = e^x
   u′ = 3x², v′ = e^x
   dy/dx = (3x² · e^x − x³ · e^x) / (e^x)²
   = e^x(3x² − x³) / e^2x
   = x²(3 − x) / e^x
5. Show Solution
   y = x cos x
   Let u = x, v = cos x
   u′ = 1, v′ = −sin x
   dy/dx = 1 · cos x + x · (−sin x) = cos x − x sin x
   
   At x = π:
   dy/dx = cos(π) − π sin(π) = −1 − π × 0 = −1
6. Show Solution
   y = e^x sin x
   Let u = e^x, v = sin x
   u′ = e^x, v′ = cos x
   dy/dx = e^x sin x + e^x cos x = e^x(sin x + cos x)
   
   Set dy/dx = 0: e^x ≠ 0, so sin x + cos x = 0
   Divide by cos x: tan x = −1
   On [0, 2π]: x = 3π/4 and x = 7π/4
   Gradient is zero at x = 3π/4 and x = 7π/4.
7. Show Solution
   y = ln x / x²
   Let u = ln x, v = x²
   u′ = 1/x, v′ = 2x
   dy/dx = ((1/x) · x² − ln x · 2x) / (x²)²
   = (x − 2x ln x) / x⁴
   = x(1 − 2 ln x) / x⁴
   = (1 − 2 ln x) / x³
8. Show Solution
   y = x² e^−x
   Let u = x², v = e^−x
   u′ = 2x, v′ = −e^−x (chain rule)
   dy/dx = 2x e^−x + x² (−e^−x) = e^−x(2x − x²) = xe^−x(2 − x)
   
   Stationary points (dy/dx = 0): e^−x ≠ 0, so x = 0 or x = 2
   At x = 0: y = 0. At x = 2: y = 4e⁻² ≈ 0.541
   Stationary points: (0, 0) and (2, 4e⁻²)
9. Show Solution
   y = x e^x
   Let u = x, v = e^x
   u′ = 1, v′ = e^x
   dy/dx = e^x + x e^x = e^x(1 + x)
   
   At x = 0: y = 0 × 1 = 0. Point: (0, 0).
   Gradient = e⁰(1 + 0) = 1
   Tangent line: y − 0 = 1(x − 0)
   y = x
10. Show Solution
    f(x) = (x² + 1) / (x² − 1)
    Let u = x² + 1, v = x² − 1
    u′ = 2x, v′ = 2x
    f′(x) = [2x(x² − 1) − (x² + 1)(2x)] / (x² − 1)²
    
    Numerator = 2x(x² − 1) − 2x(x² + 1)
    = 2x[(x² − 1) − (x² + 1)]
    = 2x[−2]
    = −4x
    
    f′(x) = −4x / (x² − 1)²
    
    f′(x) = 0 when x = 0.
    f(0) = (0 + 1)/(0 − 1) = −1
    Stationary point at (0, −1)